JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Magnetic Effects of Current

When the wire forms a circle of circumference $L$, its radius is $r=L/(2\pi )$. The magnetic force per unit length on the wire is $f=IB$ (directed radially outward). For equilibrium of a small element subtending angle $d\theta$ at the center, the radial component of tension provides the centripetal force: $2T\sin (d\theta /2)=f\cdot r\cdot d\theta$. For small $d\theta$, $\sin (d\theta /2)\approx d\theta /2$, giving $T=f\cdot r=IB\cdot \frac{L}{2\pi }=\frac{IBL}{2\pi }$. The answer is (C).